A train is approaching a tunnel which is AB in length.Ther is a cat inside the tunnel which is 3/8 distance from the starting point of the tunnel.When the train whistes the cat starts to run.The train catches the cat exactly at the entry point of the tunnel.If the cat runs towards the exit,the train catches tha cat exactly at the exit point.The speed of the train is greater than the speed of the cat in what order?

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• 9 years agoHelpfull: Yes(6) No(0)
• Answer: b) 4 : 1 worked out as under.
  
  D / v1 = (3/8) x / v2 ... ( 1 )
  
  ( D + x ) / v1 = (5/8) x / v2
  => D / v1 = (5/8) x / v2 - x / v1 ... ( 2 )
  
  From equns. ( 1 ) and ( 2 ),
  
  (3/8) x /v2 = (5/8) x / v2 - x / v1
  => 3 / (8v2) = 5 (8v2) - 1/ v1
  => 1 / v1 = ( 5/8 - 3/8) / v2
  => 4v2 = v1
• 9 years agoHelpfull: Yes(4) No(0)
• when the train croses 8 parts,the cat crosses 5 parts,,,,,,,,,,,,,so 8:5=7:4=6:3=5:2=4:1
  so the answer is 4:1
• 9 years agoHelpfull: Yes(4) No(0)
• Ans : 4 : 1
  1.Time taken for x km by train = Time taken for 3/8 of AB by cat
  2.Time taken for x+8 km by train = Time taken for 5/8 of AB by cat
  eq1 => x/v1=(3/8)AB/v2
  eq2 => (AB+x)/v1=(5/8)AB/v2 => x/v1=AB((5/8v2)-1/v1)
  LHS of eq1 and eq 2 are same so RHS of 2 eqns are equal so :
  (3/8)AB/v2 = AB((5/8v2)-1/v1)
  (3/8)/v2=(5/8v2)-1/v1
  1/v1=(2/8v2)
  1/v1=1/4v2
  v1=4v2
  
• 9 years agoHelpfull: Yes(1) No(0)